Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

suppose that (X,d) is a compact metric space and $f:X\to X$ is a continuous function. Define $X_1 = f(X), X_2 = f(X_1),...,X_{i+1} = f(X_i),...$ and let $A = \bigcap_{i=1}^\infty X_i$. Is $A \subseteq F(A)$?

share|improve this question
    
What is the need for compactness in your question? –  Elias Nov 16 '12 at 16:48
add comment

2 Answers 2

Yes. In fact, slightly more general statement is true.

Let $X, Y$ be metric spaces and $f:X\rightarrow Y$ be continuous. Let $(K_n)$ be a decreasing sequence of non-empty compact subsets of $X$. Then, $$ f(\cap_{n\in\mathbb{N}}K_n)=\cap_{n\in\mathbb{N}}f(K_n). $$ Proof follows from the nested set theorem as follows.

Let us set $$ C=\cap_{n\in\mathbb{N}}K_n. $$ Evidently. $$ f(C)\subseteq \cap_{n\in\mathbb{N}}f(K_n). $$ Conversely, let $y\in \cap_{n\in\mathbb{N}}f(K_n)$. Then, $(f^{-1}({y})\cap K_n)_{n\in\mathbb{N}}$ is decreasing sequence of non-empty compact sets. Therefore, $$ \emptyset\neq\cap_{n\in\mathbb{N}}f^{-1}({y})\cap K_n=f^{-1}({y})\cap C, $$ which proves the result.

share|improve this answer
add comment

Yes, it is true that $A\subseteq f(A)$.

Let $a\in A$, so that $a\in X_i$ for every $i$. Since $X_i = f^i(X)$, for each $i$ choose an element $x_i$ of $X$ such that $f^i(x_i)=a$. Now consider the sequence $f(x_2), f^2(x_3), f^3(x_4), \ldots$, and let $x$ be the limit of one of its convergent subsequences (which exist, because $X$ is compact). Since $f^i(x_{i+1})$ is in $X_n$ for every $i\geq n$, we must have $x\in X_n$ for each $n$ because $X_n$ is closed (being the image of a compact set in a Hausdorff space). Therefore $x\in A$. On the other hand, $f(x)$ is the limit of a convergent subsequence of $\{f^{i+1}(x_{i+1})\}$, which is constantly $a$. Therefore $f(x)=a$ and $x\in A$, so $a\in f(A)$.

On the other hand, it's true without any assumptions on $X$ and $f$ that $f(A)\subseteq A$, so in fact we have $f(A)=A$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.